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A random variable X has the following probability mass function. x 1 2 3 4 5 F(x) k2 2k2 3k2 2k 3k Find P(X > 3)

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प्रश्न

A random variable X has the following probability mass function.

x 1 2 3 4 5
F(x) k2 2k2 3k2 2k 3k

Find P(X > 3)

योग
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उत्तर

P(X > 3) = P(X = 4) + P(X = 5)

= `2/6 + 3/6`

= `5/6`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Probability Distributions - Exercise 11.2 [पृष्ठ १९४]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 11 Probability Distributions
Exercise 11.2 | Q 6. (iii) | पृष्ठ १९४

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